What are the solutions to the equation 6-2x+3x^2=0 ?

The solutions to the equation 6-2x+3x^2=0 are x= 1/3+i(sqrt(17))/3 ,x= 1/3-i(sqrt(17))/3

Find the zeros of 6-2x+3x^2=0

The zeros of 6-2x+3x^2=0 are x= 1/3+i(sqrt(17))/3 ,x= 1/3-i(sqrt(17))/3

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6 - 2 x + 3 x^{2} = 0

x = \frac{1}{3} + i \frac{17}{3}, x = \frac{1}{3} - i \frac{17}{3}

Solution steps

6 - 2 x + 3 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

3 x^{2} - 2 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

Simplify

(- 2)^{2} - 4 \cdot 3 \cdot 6 : 217 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 17 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 17 i}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 17 i}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 17 i}{2 \cdot 3} : \frac{1}{3} + i \frac{17}{3}

x = \frac{- ( - 2 ) - 2 17 i}{2 \cdot 3} : \frac{1}{3} - i \frac{17}{3}

The solutions to the quadratic equation are:

x = \frac{1}{3} + i \frac{17}{3}, x = \frac{1}{3} - i \frac{17}{3}

x, x^{2} - z^{2} = 5 y

x^{2} + 3 x - 4 = 14

2 x^{2} - 9 = 7 x

m^{2} + 63 = 0

x (2 x - 6) = 0

What are the solutions to the equation 6-2x+3x^2=0 ?
The solutions to the equation 6-2x+3x^2=0 are x= 1/3+i(sqrt(17))/3 ,x= 1/3-i(sqrt(17))/3
Find the zeros of 6-2x+3x^2=0
The zeros of 6-2x+3x^2=0 are x= 1/3+i(sqrt(17))/3 ,x= 1/3-i(sqrt(17))/3